# Leetcode [No. 190] Reverse Bits (EASY) 解題心得

## 題目

https://leetcode.com/problems/reverse-bits/description/

## 思路
+ 這個解法滿簡單的，因為我們要保證他是32bit的unsigned int，所以我們要做滿32次。
+ 每次都是自己 $*2+n的最後一個bit$

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t x = 0;
        for (int i = 0 ; i < 32; i++)
        {
            x = x << 1;
            x += n % 2;
            n = n >> 1;
            // cout << x << endl;
        }
        return x;
    }
};

### 解法分析
+ time complexity: O(1)
+ space: O(1)

### 執行結果

## follow up:

> Follow up: If this function is called many times, how would you optimize it?

不太懂意思

## Second visit on 2025/05/30

第一次寫的還比這次好，真慚愧==